Question

1)ABCD is a cyclic quadrilateral. If <ACD = 40°, <ADC = 80°. Find <CBD
2) If the ratio of the radius of two cylinder of same height is 3/4. Find the ratio of their volume​

Answer 1

1) $\sf{\huge{\red{\star}}}$ Given:-

• ABCD is a cyclic quadrilateral.
• $\sf{\angle ACD = 40^\circ}$
• $\sf{\angle ADC = 80^\circ}$

$\sf{\huge{\red{\star}}}$ To Find:-

$\sf{\angle CBD}$

$\sf{\huge{\red{\star}}}$ Note:-

Refer to the attachment provided for a better idea of the scene.

$\sf{\huge{\red{\star}}}$ Solution:-

In quad.ABCD,

$\sf{\angle ACD = 40^\circ}$

$\sf{\angle ADC = 80^\circ}$

We know,

The opposite angles of a cyclic quadrilateral are supplementary.

Hence,

$\sf{\angle ADC + \angle ABC = 180^\circ}$

= $\sf{80^\circ + \angle ABC = 180^\circ}$

= $\sf{\angle ABC = 180^\circ - 80^\circ}$

= $\sf{\angle ABC = 100^\circ}$

Now,

We also know,

Angles on the same segment of a cyclic quadrilateral are always equal.

Hence,

$\sf{\angle ABD = \angle ACD}$

=> $\sf{\angle ABD = 40^\circ}$

Now,

$\sf{\angle ABD + \angle CBD = \angle ABC}$

= $\sf{40^\circ + \angle CBD = 100^\circ}$

= $\sf{\angle CBD = 100^\circ - 40^\circ}$

= $\sf{\angle CBD = 60^\circ}$

$\sf{\bold{\therefore\:\angle CBD = 60^\circ}}$

______________________________________

2. $\sf{\huge{\red{\star}}}$ Given:-

• Ratio of radius of two cylinders = 3:4
• Height of both the cylinders are same.

$\sf{\huge{\red{\star}}}$ To Find:-

Ratio of volume of the two cylinders.

$\sf{\huge{\red{\star}}}$ Assumption:-

Let the radius of 1st cylinder be $\sf{r_1}$ and the radius of 2nd cylinder be $\sf{r_2}$

Let the height of both the cylinders be h.

$\sf{\huge{\red{\star}}}$ Solution:-

We know,

Volume of a cylinder = πr²h

Therefore for 1st cylinder,

$\sf{Volume\:(V_1) = \pi (r_1)^2 h}$

= $\sf{V_1 = \pi (3)^2 h}$

= $\sf{V_1 = \pi 9 h}$

Now,

For the 2nd cylinder,

$\sf{Volume\:(V_2) = \pi (r_2)^2 h}$

= $\sf{V_2 = \pi (4)^2 h}$

= $\sf{V_2 = \pi 16 h}$

Now,

Ratio of the Volumes of both the cylinder =

$\sf{Volume\:of\:1st\:cylinder\:(V_1):Volume\:of\:2nd\:cylinder\:(V_2)}$

= $\sf{\pi 9h : \pi 16h}$

= $\sf{\dfrac{\pi 9 h}{\pi 16h}}$

= $\sf{\dfrac{\cancel{\pi} 9 \cancel{h}}{\cancel{\pi} 16 \cancel{h}}}$

= $\sf{\dfrac{9}{16}}$

= $\sf{9:16}$

Therefore the ratio of volumes of both the cylinders is 9:16.

______________________________________

Answer 2

$\large{\underline{\purple{\frak{ \: Given:- \: }}}}$

• $\sf \angle ACD = 40^{\circ}$
• $\sf \angle ADC = 80^{\circ}$

$\large{\underline{\purple{\frak{ \: Find:- \: }}}}$

• $\sf Value\:of\:\angle CBD$

$\large{\underline{\purple{\frak{ \: Diagram:- \: }}}}$

$\setlength{\unitlength}{1cm}\begin{picture}(0,0)\thicklines\qbezier(2.3,0)(2.121,2.121)(0,2.3)\qbezier(-2.3,0)(-2.121,2.121)(0,2.3)\qbezier(-2.3,0)(-2.121,-2.121)(0,-2.3)\qbezier(2.3,0)(2.121,-2.121)(-0,-2.3) \qbezier(1.7,1.6)(1.7,2)(1.7,-1.6)\qbezier(-1.5,1.7)(-1.5,2)( -1.5,-1.7)\qbezier( - 1.5,1.7)(1.7,1.6)(1.7,1.6) \qbezier(-1.6, - 1.7)(1.7,-1.6)(1.7,-1.6)\qbezier( - 1.5,1.7)(1.7,-1.6)(1.7,-1.6)\qbezier(-1.6, -1.7)(1.7,1.6)(1.7,1.6)\qbezier(1.1, -1)(1,-1.5)(1,-1.6)\qbezier(-1,-1.7)( - 0.8,-0.7)(-1.5,-1)\put(-2,1.8){\bf A}\put(1.7,1.7){\bf B} \put(1.7, -1.9){\bf C} \put( - 2, - 1.9){\bf D}\put(0.43,- 1.4){ \bf {40}^{\circ} }\put( -1.2,-0.7){ \bf {80}^{\circ} } \end{picture}$

$\large{\underline{\purple{\frak{ \: Solution:- \: }}}}$

Let us f1st recall the properties of cyclic quadrilateral. In a cyclic quadrilateral sum of opp. angles is 180° and in a Circle angles made from the same segments are equal.

Here,

$\sf \implies \angle ACD = \angle ABD = 40^{\circ} \qquad \bigg \lgroup Angles\:in\:same\: segment \bigg \rgroup$

$\sf \implies \angle ABC + \angle ADC = 180^{\circ} \qquad \bigg \lgroup Opp. \:angles\:of\: cyclic\:quadr. \bigg \rgroup$

$\sf where \footnotesize{\begin{cases} \\ \sf\angle ADC = 80^{\circ} \\ \\ \end{cases}}$

Substitute this value:-

$\sf \implies \angle ABC + \angle ADC = 180^{\circ}$

$\sf \implies \angle ABC + 80^{\circ} = 180^{\circ}$

$\sf \implies \angle ABC= 180^{\circ} - 80^{\circ}$

$\sf \implies \angle ABC=100^{\circ}$

Now,

$\sf \dashrightarrow \angle ABC = \angle ABD + \angle CBD$

$\sf \dashrightarrow \angle ABC - \angle ABD = \angle CBD$

$\sf where \footnotesize{\begin{cases}\sf\angle ABC = 100^{\circ}\\ \sf\angle ABD = 40^{\circ} \end{cases}}$

Substituting these values:-

$\sf \dashrightarrow \angle CBD = \angle ABC - \angle ABD$

$\sf \dashrightarrow \angle CBD = 100^{\circ} - 40^{\circ}$

$\sf \dashrightarrow \angle CBD = 60^{\circ}$

$\underline{\boxed{\sf \therefore Value \:of\:\angle CBD=60^{\circ}}}$

____________________________

$\large{\underline{\pink{\frak{ \: Given:- \: }}}}$

• $\textsf{Ratio of radius of two cylinder is 3:4}$

$\large{\underline{\pink{\frak{ \: Find:- \: }}}}$

• $\textsf{Ratio of their volume.}$

$\large{\underline{\pink{\frak{ \: Solution:- \: }}}}$

Let, Radius of 1st cylinder be 'R' = 3

Radius of 2nd Cylinder be 'r' = 4

Height of both cylinder is 'h'

Now, using

$\huge{\underline{\boxed{ \sf Volume \: of \: cylinder = \pi {r}^{2}h }}}$

$\underline{\red{\textsf{For 1st cylinder}}}$

$\implies\sf Volume \:_{1st \: cylinder} = \pi {R}^{2}h$

$\implies\sf Volume \:_{1st \: cylinder} = \pi {(3)}^{2}h$

$\implies\sf Volume \:_{1st \: cylinder} = \pi 9h$

$\implies\sf Volume \:_{1st \: cylinder} = 9\pi h$

$\underline{\purple{\textsf{For 2nd cylinder}}}$

$\implies\sf Volume \:_{2nd \: cylinder} = \pi {r}^{2}h$

$\implies\sf Volume \:_{2nd \: cylinder} = \pi {(4)}^{2}h$

$\implies\sf Volume \:_{2nd \: cylinder} = \pi 16h$

$\implies\sf Volume \:_{2nd \: cylinder} = 16\pi h$

$\quad\qquad$ ________________

$\dashrightarrow\sf Ratio_{volumes}= \dfrac{Volume \:_{1st \: cylinder}}{Volume \:_{2nd \: cylinder}}$

$\dashrightarrow\sf Ratio_{volumes}= \dfrac{9 \pi h}{16 \pi h }$

$\dashrightarrow\sf Ratio_{volumes}= \dfrac{9 \not\pi \not h}{16 \not\pi \not h }$

$\dashrightarrow\sf Ratio_{volumes}= \dfrac{9}{16}$

$\dashrightarrow\sf Ratio_{volumes}= 9:16$

$\underline{\boxed{\sf \therefore Ratio\:of\:their\: volumes\:is\:9:16}}$